Categories: Mathematics, Physics.

Fundamental solution

Given a linear operator L^\hat{L} acting on x[a,b]x \in [a, b], its fundamental solution G(x,x)G(x, x') is defined as the response of L^\hat{L} to a Dirac delta function δ(xx)\delta(x - x') for x]a,b[x \in ]a, b[:

L^{G(x,x)}=Aδ(xx)\begin{aligned} \boxed{ \hat{L}\{ G(x, x') \} = A \delta(x - x') } \end{aligned}

Where AA is a constant, usually 11. Fundamental solutions are often called Green’s functions, but are distinct from the (somewhat related) Green’s functions in many-body quantum theory.

Note that the definition of G(x,x)G(x, x') generalizes that of the impulse response. And likewise, due to the superposition principle, once GG is known, L^\hat{L}’s response u(x)u(x) to any forcing function f(x)f(x) can easily be found as follows:

L^{u(x)}=f(x)    u(x)=1Aabf(x)G(x,x)dx\begin{aligned} \hat{L} \{ u(x) \} = f(x) \quad \implies \quad \boxed{ u(x) = \frac{1}{A} \int_a^b f(x') \: G(x, x') \dd{x'} } \end{aligned}

L^\hat{L} only acts on xx, so x]a,b[x' \in ]a, b[ is simply a parameter, meaning we are free to multiply the definition of GG by the constant f(x)f(x') on both sides, and exploit L^\hat{L}’s linearity:

Af(x)δ(xx)=f(x)L^{G(x,x)}=L^{f(x)G(x,x)}\begin{aligned} A f(x') \: \delta(x - x') = f(x') \hat{L}\{ G(x, x') \} = \hat{L}\{ f(x') \: G(x, x') \} \end{aligned}

We then integrate both sides over xx' in the interval [a,b][a, b], allowing us to consume δ(x ⁣ ⁣x)\delta(x \!-\! x'). Note that dx\int \dd{x'} commutes with L^\hat{L} acting on xx:

Aabf(x)δ(xx)dx=abL^{f(x)G(x,x)}dxAf(x)=L^abf(x)G(x,x)dx\begin{aligned} A \int_a^b f(x') \: \delta(x - x') \dd{x'} &= \int_a^b \hat{L}\{ f(x') \: G(x, x') \} \dd{x'} \\ A f(x) &= \hat{L} \int_a^b f(x') \: G(x, x') \dd{x'} \end{aligned}

By definition, L^\hat{L}’s response u(x)u(x) to f(x)f(x) satisfies L^{u(x)}=f(x)\hat{L}\{ u(x) \} = f(x), recognizable here.

While the impulse response is typically used for initial value problems, the fundamental solution GG is used for boundary value problems. Suppose those boundary conditions are homogeneous, i.e. u(x)u(x) or one of its derivatives is zero at the boundaries. Then:

0=u(a)=1Aabf(x)G(a,x)dx    G(a,x)=00=ux(a)=1Aabf(x)Gx(a,x)dx    Gx(a,x)=0\begin{aligned} 0 &= u(a) = \frac{1}{A} \int_a^b f(x') \: G(a, x') \dd{x'} \qquad \implies \quad G(a, x') = 0 \\ 0 &= u_x(a) = \frac{1}{A} \int_a^b f(x') \: G_x(a, x') \dd{x'} \quad \implies \quad G_x(a, x') = 0 \end{aligned}

This holds for all xx', and analogously for the other boundary x=bx = b. In other words, the boundary conditions are built into GG.

What if the boundary conditions are inhomogeneous? No problem: thanks to the linearity of L^\hat{L}, those conditions can be given to the homogeneous solution uh(x)u_h(x), where L^{uh(x)}=0\hat{L}\{ u_h(x) \} = 0, such that the inhomogeneous solution ui(x)=u(x)uh(x)u_i(x) = u(x) - u_h(x) has homogeneous boundaries again, so we can use GG as usual to find ui(x)u_i(x), and then just add uh(x)u_h(x).

If L^\hat{L} is self-adjoint (see e.g. Sturm-Liouville theory), then the fundamental solution G(x,x)G(x, x') has the following reciprocity boundary condition:

G(x,x)=G(x,x)\begin{aligned} \boxed{ G(x, x') = G^*(x', x) } \end{aligned}

Consider two parameters x1x_1' and x2x_2'. The self-adjointness of L^\hat{L} means that:

abG(x,x1)(L^{G(x,x2)})dx=ab(L^{G(x,x1)})G(x,x2)dxabG(x,x1)δ(xx2)dx=abδ(xx1)G(x,x2)dxG(x2,x1)=G(x1,x2)\begin{aligned} \int_a^b G^*(x, x_1') \Big( \hat{L} \{ G(x, x_2') \} \Big) \dd{x} &= \int_a^b \Big( \hat{L} \{ G(x, x_1') \} \Big)^* G(x, x_2') \dd{x} \\ \int_a^b G^*(x, x_1') \: \delta(x - x_2') \dd{x} &= \int_a^b \delta^*(x - x_1') \: G(x, x_2') \dd{x} \\ G^*(x_2', x_1') &= G(x_1', x_2') \end{aligned}


  1. O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.